# Number of subarrays with sum greater than k

Lets take an array of size n.There 2 n possible **subarrays** **of** this array. Lets take the example array of size 4: [1, 2, 3, 4].There are 2 4 **sub** **arrays** .. **Sub** **array** **of** the empty set ([]) is the 0 th one (0000).The **subarray** **of** [1], is the second one (0001), the **subarray** **of** [2] is the second one... (0010) and the **subarray** [1, 2] is the third one. Max Array **Sum** , is a HackerRank problem from Dynamic Programming subdomain Given a 2D Array, : 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 We define an hourglass in to be a subset of so what we. neenah homes for. Given an integer array nums and an integer **k**, return the length of the shortest non-empty subarray of nums with a **sum** of at least **k**. If there is no such subarray, return -1. A subarray is a contiguous part of an array. Example 1: Input: nums = [1], **k** = 1 Output: 1 Example 2: Input: nums = [1,2], **k** = 4 Output:-1 Example 3: Input: nums = [2,-1,2. which **numbers** are **greater** or lesser. Negative **Numbers** (−) Positive **Numbers** (+) (The line continues left and right forever.) A **number** on the left is less than a **number** on the right. Examples: 5 is less than 8; −1 is less than 1; −8 is less than −5; A **number** on the right is **greater** **than** a **number** on the left. **Subarray** with 0 **sum** brute force. private pet friendly rentals; capital one developer academy associate salary.

lp

Base case conditions: If the target **sum** is zero or less than zero, return 0. If the **sum** **of** the array is smaller than the target **sum**, return -1. If the **sum** **of** the array is equal to the target **sum**, return the length of the array. If the first element in the array is **greater** **than** one, return 1. (First element in the array is the smaller subset. Maximize count of non-overlapping **subarrays** **with** **sum** **K** . Given an array arr[] consisting of N integers and an integer X, the task is to count the maximum **number** **of** subsets possible from the given array having. Smallest element of the Subset * Size of the Subset ≥ X. Examples: _Input: _arr[] = {7, 11, 2, 9, 5}, X = 10 Output:_ 2_ _Explanation: _. 1992 f150 freon capacity international 345 vs chevy 350 bmw m57 vacuum pump failure hudspeth county. Find **subarrays** of an array; kawasaki fx691v problems; things to do near guaranteed rate field; why is life 360 not updating. Given an array and a desired **sum** (red cells define a subarray **with sum** equal to ): As we can see, the answer here is four because there are **subarrays** with a **sum** equal to . 3. Naive Approach. 3.1. Main Idea. The main idea in this approach is to check for each possible subarray if its **sum** is equal to , or not. Given an array of positive integers nums and a positive integer target, return the minimal length of a contiguous subarray [nums l, nums l+1, ..., nums r-1, nums r] of which the **sum** is **greater than** or equal to target. If there is no such subarray, return 0 instead. Example 1:. "/>. The **sum** of an array is the **sum** of its elements. Given an element array of integers, , and an integer, , determine the maximum value of the **sum** of any of its **subarrays** modulo .. Input the array size, window size and the array elements. If the array size is **greater than** the window size, then call the function to find the maximum **sum**. The idea is to generate all the **subarrays** **of** the given array and check whether **sum** **of** elements of the **subarray** is equal to given **k** . If **sum** **of** the **subarray** elements is equal to given **k** then increment the value of count used to store the required result. Algorithm. Run outer loop from range[0 to n-1]. If it were 0, we'd be done, but it isn't. Now, we start with the largest **subarrays**, length 6. We note that the first **subarray** is the same as the whole array, except for the last element; so, the **sum** is **sum** 0 0 minus the last element, which is -3. The **sum** is -4 - -3 = -1. This is **sum** 1 0. Dismiss. . The function should find and return the count of total **number** of continuous **subarrays** whose **sum** equals to the **number** specified by the second argument. It is guaranteed that all the **numbers** in the array are positive **numbers**. For example −. If the inputs are −. const arr = [1, 1, 1]; const **sum** = 2;. Initialise **sum** to 0. Write a loop that iterates till n. Add current element to **sum**. If the **sum** is **greater** **than** **k**. The maximum **subarray** length is i + 1. Else the maximum **subarray** length is. Find the index from the previous **sums** using binary search. The **sum** that is less than **sum** - **k** - 1 is the element index we want.

ni

cj

rf

xj

vn

xg

Problem Given an array, count the total **number** **of** strictly increasing **subarrays** in it. Sample Input [1, 2, 3, 1] Sample Output 3 Explanation Let's say the endpoints of a strictly increasing **subarray** are start and end . Then, the **subarray** arr[start, end+1] will be strictly increasing if the element at end+1 is **greater** **than** the element at end . The same thing goes for elements at end+1, end+2. aria-label="Show more">. In this article, we will find out the **number** of **subarrays** having a **sum** less **than K** using C++. In this problem, we have an array arr [] and an integer **K**. So now we have to find **subarrays** that have a **sum** less **than K**. Here is the example −. Input : arr [] = {1, 11, 2, 3, 15} **K** = 10 Output : 4 {1}, {2}, {3} and {2, 3}.

zx

ca

. Simple Approach: The simple approach to solve this problem is to run two for loops and for every **subarray** check if it is the maximum **sum** possible. Follow the below steps to solve the problem. Run a loop for i from 0 to n - 1, where n is the size of the array. Now, we will run a nested loop for j from i to n - 1 and add the value of the. There are as many such arrays as the length of the window. 2nd case : If **sum** becomes **greater** **than** or equal to **k**, this means we need to subtract starting element from **sum** so that the **sum** again becomes less than **k**. So we adjust the window's left border by incrementing start. We follow the same procedure until end < array size. Implementation: C++. Given an array of integers arr, find the **sum** of min(b), where b ranges over every (contiguous) **subarray** of arr. Since the answer may be large, return the answer modulo 109 + 7 . Example 1:. "/> panzer m4 shotgun price near seoul; honda pioneer price; outlook vba macro.

qc

what are the 66 books of the bible in order. remarkably bright creatures movie. frontier soda alaska. To **sum** if **greater** **than**, you can use the SUMIF function. In the example shown, cell H6 contains this formula: = SUMIF( amount,">1000") where "amount" is a named range for cells D5:D11. This formula **sums** the amounts in column D when they are **greater** **than** 1000. The **sum** **of** an array is the **sum** **of** its elements. Given an element array of integers, , and an integer, , determine the maximum value of the **sum** **of** any of its **subarrays** modulo .. Input the array size, window size and the array elements. If the array size is **greater** **than** the window size, then call the function to find the maximum **sum**. From the map find **number** of **subarrays** previously found having **sum** equal to currsum-**sum**. Excluding all those **subarrays** from current subarray, gives new **subarrays** having desired **sum**. So increase count by the **number** of such **subarrays**. Note that when currsum is equal to desired **sum** then also check **number** of **subarrays** previously having **sum** equal to. Step 1: Find **sum** **of** first **k** elements in the input array. Initialize maxSum to the calculated **sum** and maxSumStartIndex = 0. Step 2: Add next element to the **sum** and subtract first element from the **sum**. Check if this **sum** is **greater** **than** previous **sum** and update maxSum and maxSumStartIndex. Step 3: Keep adding next element to the **sum** and removing. Approach: The idea is to store the **sum** **of** elements of every prefix of the array in a hashmap, i.e, every index stores the **sum** **of** elements up to that index hashmap. So to check if there is a **subarray** **with** a **sum** equal to s, check for every index i, and **sum** up to that index as x.If there is a prefix with a **sum** equal to x - s, then the **subarray** **with** the given **sum** is found. In this article, we will find out the **number** of **subarrays** having a **sum** less **than K** using C++. In this problem, we have an array arr [] and an integer **K**. So now we have to find **subarrays** that have a **sum** less **than K**. Here is the example −. Input : arr [] = {1, 11, 2, 3, 15} **K** = 10 Output : 4 {1}, {2}, {3} and {2, 3}. This video explains a very important programming interview problem which is to count the **number** **of** **subarrays** in a given array with **sum** exactly equals to **K**. T. 2. If the sum_here is less than X then add the element pointed by j to variable sum_here and increment j. 3. If the sum_here is **greater** **than** X then decrement sum_here by the value pointed by i and increment i. 4. If the sum_here equals to X then we got the **subarray** starting with i and ending point is j. 5. If the sum_here is **greater** **than** X then decrement sum_here by the value pointed by i and increment i. 4. ... Count **subarrays** **with** equal **number** **of** 1's and 0's: 221: 317: Minimum swaps required to bring all elements less than or equal to **k** together: 220: 318: Reverse individual words: 218: 319:.

Problem Statement: Given an array and a **sum k**, we need to print the length of the longest subarray that sums to **k**. Examples: Example 1: Input: arr = {7,1,6,0}, **k** = 7 Output: Length of the longest subarray **with sum K** is 3 Explanation: 1 + 6 + 0 = 7, it is the longest subarray **with sum** 7 and length 3.Example 2: Input: arr = {2,3,5,1,9}, **k** = 10 Output: Length of the longest. 1.2 Smallest **Subarray** with a given **sum** (easy) Problem Statement. Given an array of positive **numbers** and a positive **number** 'S,' find the length of the smallest contiguous **subarray** whose **sum** is **greater than** or equal to 'S'. Return 0 if no such **subarray** exists. Example 1:. Max Array **Sum** , is a HackerRank problem from Dynamic Programming subdomain Given a 2D Array, : 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 We define an hourglass in to be a subset of so what we. neenah homes for.

cc

Approach 1: Brute Force. A simple solution could be to traverse all the **subarrays** and calculate their **sum**. If the **sum** is equal to the given required **sum**, then increment the count of **subarrays**. Finally, return the count of such **subarrays**. The algorithm to calculate the required will be:-. Initialize a variable ‘res’ as 0 to store the count. Detailed solution for **Subarray** **with** Given **Sum** - Problem Statement: **Subarray** **with** Given **Sum** Given an array and a **sum** **k**, generate the **subarray** whose elements **sum** to **k**. Examples: Example 1: Input: arr = {1, 7, 3, 9}, **k** = 10 Output: 7 3 Explanation: Of all the **subarrays**, 7 and 3 **sums** to 10. Example 2: Input: arr = {2,1,3,4,5,6}, **k**. If it were 0, we'd be done, but it isn't. Now, we start with the largest **subarrays**, length 6. We note that the first subarray is the same as the whole array, except for the last element; so, the **sum** is **sum** 0 0 minus the last element, which is -3. The **sum** is -4 - -3 = -1. This is **sum** 1 0. Dismiss. Problem Given an array, count the total **number** **of** strictly increasing **subarrays** in it. Sample Input [1, 2, 3, 1] Sample Output 3 Explanation Let's say the endpoints of a strictly increasing **subarray** are start and end . Then, the **subarray** arr[start, end+1] will be strictly increasing if the element at end+1 is **greater** **than** the element at end . The same thing goes for elements at end+1, end+2. From the map find **number** of **subarrays** previously found having **sum** equal to currsum-**sum**. Excluding all those **subarrays** from current subarray, gives new **subarrays** having desired **sum**. So increase count by the **number** of such **subarrays**. Note that when currsum is equal to desired **sum** then also check **number** of **subarrays** previously having **sum** equal to. 2. If the **sum**_here is less **than** X then add the element pointed by j to variable **sum**_here and increment j. 3. If the **sum**_here is **greater than** X then decrement **sum**_here by the value pointed by i and increment i. 4. If the **sum**_here equals to X then we got the subarray starting with i and ending point is j. 5. The **numbers** present from the 1st to 3rd indexes are 2, 4, 7. When we add (2 + 4 + 7) it is 13. ... We run two for loops to make all possible **subarrays** **sum**. In the inner loop, we check if the value of the **sum** is equal to **k**. ... If the value of **sum** is **greater** **than** **k** then move the start pointer and also subtract the value present at the start. Lets take an array of size n.There 2 n possible **subarrays** **of** this array. Lets take the example array of size 4: [1, 2, 3, 4].There are 2 4 **sub** **arrays** .. **Sub** **array** **of** the empty set ([]) is the 0 th one (0000).The **subarray** **of** [1], is the second one (0001), the **subarray** **of** [2] is the second one... (0010) and the **subarray** [1, 2] is the third one.

xx

lo

Largest **number** less **than** N with digit **sum greater than** the digit **sum** of N. 04, Dec 18. **Number** of **subarrays** having even product. 24, Feb 21. Count **subarrays** having product equal to the power of a given Prime **Number** . 22, Mar 21. **Number** of **subarrays** having absolute **sum greater than K** | Set-2. 03, Jul 19. **Numbers** less **than** N which are product of exactly two. Given an array of integers arr and two integers **k** and threshold. Return the **number** **of** **sub-arrays** **of** size **k** and average **greater** **than** or equal to threshold. Example 1: Input: arr = [2,2,2,2,5,5,5,8], **k** = 3, threshold = 4 Output: 3 Explanation: **Sub-arrays** [2,5,5],[5,5,5] and [5,5,8] have averages 4, 5 and 6 respectively. All other **sub-arrays** **of**. tabindex="0" title=Explore this page aria-label="Show more">. The task is to return the **number** **of** **subarrays** having value strictly greate. Problems ... **Subarray** **with** given **sum**. Check for BST. Minimum **number** **of** jumps. Kadane's Algorithm ... 3 Explanation: The **subarrays** having value strictly **greater** **than** **K** are: [3], [3, 2] and [3, 2, 1]. Thus there are 3 such **subarrays**. Example 2: Input: N = 4, **K** = 1 Arr.

Given a positive integer array nums and an integer **k**, return the **number** of non-empty **subarrays** of nums whose score is strictly less **than k**. A subarray is a contiguous sequence of elements within an array. Example 1: Input: nums = [2,1,4,3,5], **k** = 10 Output: 6 Explanation: The 6 **subarrays** having scores less **than** 10 are: - [2] with score 2 * 1 = 2. There is no special intuition needed for the brute force solution. We just need to generate all possible **subarrays** and check the **sum** . Approach: In order to generate all the **subarrays** , we need two nested loops. First loop decides the starting index of the **subarray**. Record the max **sum** as 0. If current **sum** is **greater** **than** the max **sum**, assign it to the max **sum**. If current **sum** is less than 0, restart summing from the next element. Time complexity: O(n) (linear time complexity) Space complexity: O(1) (constant space complexity) Inverse the "less than / **greater** **than**" checks to find the minimum **subarray** **sum**. Suppose I have a vector containing n elements. I want to find out the **number** **of** previous elements **greater** **than** its element at present index i. I want to find A[i] > A[j] && i < j for all elements of the vector. Constraints: 1 <= t <= 1000 1 <= n <= 10^5 1 <= A[i] <=10^5 My code:.

he

. Leetcode 1292. Maximum Side Length of a Square with **Sum** Less than or Equal to Threshold (09 Apr 2022) Leetcode 1314. Matrix Block **Sum** (09 Apr 2022) Leetcode 1343. **Number** **of** **Sub-arrays** **of** Size **K** and Average **Greater** **than** or Equal to Threshold (09 Apr 2022) Leetcode 1371. Find the Longest Substring Containing Vowels in Even Counts (09 Apr 2022. A **subarray** is a contiguous sequence of elements within an array. Example 1: Input: nums = [2,1,4,3,5], **k** = 10 Output: 6 Explanation: The 6 **subarrays** having scores less than 10 are: - [2] with score 2 * 1 = 2. - [1] with score 1 * 1 = 1. - [4] with score 4 * 1 = 4. - [3] with score 3 * 1 = 3. Max **Sum** **of** Rectangle No Larger Than **K**; 365. Water and Jug Problem; 368. Largest Divisible Subset; 371. **Sum** **of** Two Integers ... Next **Greater** Node In Linked List; 1020. **Number** **of** Enclaves; 1021. Remove Outermost Parentheses; ... **Sum** **of** All Odd Length **Subarrays**; 1598. Crawler Log Folder; 1599. Maximum Profit of Operating a Centennial Wheel.

If the sum_here is **greater** **than** X then decrement sum_here by the value pointed by i and increment i. 4. ... Count **subarrays** **with** equal **number** **of** 1's and 0's: 221: 317: Minimum swaps required to bring all elements less than or equal to **k** together: 220: 318: Reverse individual words: 218: 319:. Because there are as many subarrays as the length of the window. If the sum is greater than or equal to K, we need to subtract the starting element from the sum so that it becomes less than K again and increment our start pointer by 1. Keep repeating steps 1 and 2 until we reach the end of the array. ILLUSTRATION:. First note that the answer will always use min(k - n, 0) **subarrays** **with** maximal **sums**. Sof let us find the **sum** **of** min(k - n, 0) maximal **subarrays**, elements that are used in them, and the following min(k, n) **subarrays**. We can do it using binary search for the border **sum**, and a data structure similar to Fenwick tree. The idea is to generate all the **subarrays** **of** the given array and check whether **sum** **of** elements of the **subarray** is equal to given **k** . If **sum** **of** the **subarray** elements is equal to given **k** then increment the value of count used to store the required result. Algorithm. Run outer loop from range[0 to n-1]. PYTHON Program to find the **number** **of** **subarrays** **with** product less than a value **K** Naive Approach : Key Idea:First we check for each value if it is less than K.Then we will check for all other **subarray** multiplications with that value.If it is less than **K** then we increase counter by 1 else we break as every next product will be even more.

zw

If a **sub-array** **sums** up to **k**, then the **sum** at the end of this **sub-array** will be sumEnd = sumStart + **k**. That implies: sumStart = sumEnd - **k**. Suppose, at index 10, **sum** = 50, and the next 6 **numbers** are 8,-5,-3,10,15,1. At index 13, **sum** will be 50 again (the **numbers** from indexes 11 to 13 add up to 0). Then at index 16, **sum** = 76. Given an unsorted array of integers, find the **number** **of** **subarrays** having **sum** exactly equal to a given **number** **k**. Examples: Input : arr [] = {10, 2, -2, -20, 10}, **k** = -10 Output : 3 **Subarrays**: arr [0...3], arr [1...4], arr [3..4] have **sum** exactly equal to -10. Lets take an array of size n.There 2 n possible **subarrays** **of** this array. Lets take the example array of size 4: [1, 2, 3, 4].There are 2 4 **sub** **arrays** .. **Sub** **array** **of** the empty set ([]) is the 0 th one (0000).The **subarray** **of** [1], is the second one (0001), the **subarray** **of** [2] is the second one... (0010) and the **subarray** [1, 2] is the third one. There are two **subarrays** whose **sum** is 2: First from index 0 to 1 i.e. [1, 1], Second from index 1 to 2 i.e. [1, 1] Approach Idea: The main idea is to use a hash map to store the frequency of prefix sums. So, we’ll iterate over the array, finding how many **subarrays** exist whose **sum** equals **k** and ends at the current point on each iteration. If the array size is **greater** **than** the window size, then call the function to find the maximum **sum**. Use the first loop to iterate the k-sized loop to find the window **sum** and after each complete iteration of the second loop (window sized loop) check for the max **sum**. Finally, print the maximum **sum** value with its window size.

da

tu

1992 f150 freon capacity international 345 vs chevy 350 bmw m57 vacuum pump failure hudspeth county. Find **subarrays** of an array; kawasaki fx691v problems; things to do near guaranteed rate field; why is life 360 not updating. The idea is to maintain a window that ends at the current element, and the **sum** **of** its elements is less than or equal to the given **sum**. If the current window's **sum** becomes more than the given **sum** at any point of time, then the window is unstable and continue removing elements from the window' left till it becomes stable again. On entering the 'for' loop, check whether the ith element of the input array is zero or not. And if it is a zero then put '-1' at array [i]. Then add the ith element of array to **'sum'**. After that check if the **'sum'** equals zero or not. If yes then update 'maxlen' with 'i+1'. But if the **'sum'** is not equal to '0' then check if the map contains. subarray only uses that part of awhere the elements of the subarray are located. For example, if aleft occupies locations 0 through 10 of a, then only those locations will be a ected when Quicksort is called on input aleft. It is this in-place property that gives Quicksort an. blackhead popping videos 2022; fn fal vs m14; carshield commercial actress julie; stock turnover ratio is. Line 27 checks if the **sum** **of** the current **subarray** is **greater** **than** the maximum **sum** being maintained so far. If yes, it updates the result (line 29)and saves the resultant **subarray's** starting index. This approach takes O(n 3) time as the **subarray** **sum** is calculated in O(1) time for each of n 2 **subarrays** **of** an array of size n, and it takes O(n) time to print a **subarray**.. 2. Hashing. We can also use hashing to find **subarrays** **with** the given **sum** in an array by using a map of lists or a multimap for storing the end index of all **subarrays** having a given **sum**. The idea is to traverse the given.

ux

id

hs

nr

wh

Output − Count of subarrays with all elements greater than K are − 10 Explanation − Subaarays will be − [3], [4], [5], [6], [3,4], [4,5], [5,6], [3,4,5], [4,5,6], [3,4,5,6]. Total count=10. The approach used in the below program is as follows We will traverse the array using a for a loop. If the current element is greater than K. Increment count. Last Edit: July 3, 2021 5:34 PM. Read More. Problem: Given an array of n integer **numbers**, find maximum **sum** **of** **sub** **array** which has **sum** less than or equal to **K**. Solution: class Solution { long findMaxSubarraySum(long arr [], int n, int **k**) { long right = 0; TreeSet<Long> bst = new TreeSet<> (); bst.add (0L); long ans = Long.MIN_VALUE; for (int i. Max Array **Sum** , is a HackerRank problem from Dynamic Programming subdomain Given a 2D Array, : 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 We define an hourglass in to be a subset of so what we. neenah homes for. Simple Approach: The simple approach to solve this problem is to run two for loops and for every **subarray** check if it is the maximum **sum** possible. Follow the below steps to solve the problem. Run a loop for i from 0 to n - 1, where n is the size of the array. Now, we will run a nested loop for j from i to n - 1 and add the value of the. Example: maximum element in a window of size **k** #include <bits/stdc++.h> using namespace std; // A Dequeue (Double ended queue) based method for printing maximum element of // all **subarrays** **of** size **k** void printKMax(int arr[], int n, int **k** ) { // Create a Double Ended Queue, Qi that will store indexes of **array** elements // The queue will store indexes of useful elements in every window and. Unlike subsequences, **subarrays** are required to occupy consecutive positions within the original array. We can solve this problem by using the sliding window technique. The idea is to maintain a window of size **k**. For every array element, include it in the window and remove the window's leftmost element if the window size is more than **k**. Also.

bm

vg

Given an array of real **numbers**** greater than** zero in form of strings. Find if there exists a triplet (a,b,c) such that 1 < a+b+c < 2 . Return 1 for true or 0 for false. The horizontal axis is the index **k**, the **number** of occurrences. ... the **number** of meteorites **greater than** 1 meter diameter that strike Earth in a year; the **number** of patients arriving in an emergency room between 10 and 11 pm; and ; the **number** of laser photons hitting a detector in a particular time interval. Assumptions and validity. The **Poisson distribution** is an appropriate model if the. The first line of each test case contains one integer n ( 1 ≤ n ≤ 10 5 ) — the length of the array a. The second line of each test case contains a string consisting of n decimal digits, where the i -th digit is equal to the value of a i. It is guaranteed that the **sum** **of** n over all test cases does not exceed 10 5. Solution 2: Lets say array is arr [] and given **sum** is X. Iterate over array arr []. If currentSum is less than X then add current element to currentSum. If currentSum is **greater** **than** X , it means we need to remove starting elements to make currentSum less than X. If CurrentSum is equal to X, we got the continuous **sub** **array,** print it. Approach: The idea is to store the **sum** **of** elements of every prefix of the array in a hashmap, i.e, every index stores the **sum** **of** elements up to that index hashmap. So to check if there is a **subarray** **with** a **sum** equal to s, check for every index i, and **sum** up to that index as x.If there is a prefix with a **sum** equal to x - s, then the **subarray** **with** the given **sum** is found. Learn to code finding the subarray with maximum **sum** using divide and conquer . Covering algorithms operate by adding tests to the rule that is under construction, always striving to create a rule with maximum accuracy. 1 5 4 8 10 2 6 9 12 11 3 7 1 5 4 8 10 2 6 9 12 11 3 7 Divide : O(1). A Computer Science portal for geeks. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions.

ok

qn

A Computer Science portal for geeks. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. aria-label="Show more">. Given an integer array nums and an integer **k**, return the length of the shortest non-empty subarray of nums with a **sum** of at least **k**. If there is no such subarray, return -1. A subarray is a contiguous part of an array. Example 1: Input: nums = [1], **k** = 1 Output: 1 Example 2: Input: nums = [1,2], **k** = 4 Output:-1 Example 3: Input: nums = [2,-1,2. Because there are as many subarrays as the length of the window. If the sum is greater than or equal to K, we need to subtract the starting element from the sum so that it becomes less than K again and increment our start pointer by 1. Keep repeating steps 1 and 2 until we reach the end of the array. ILLUSTRATION:. Learn to code finding the subarray with maximum **sum** using divide and conquer . Covering algorithms operate by adding tests to the rule that is under construction, always striving to create a rule with maximum accuracy. 1 5 4 8 10 2 6 9 12 11 3 7 1 5 4 8 10 2 6 9 12 11 3 7 Divide : O(1). .

bv

yw

Largest **number** less **than** N with digit **sum greater than** the digit **sum** of N. 04, Dec 18. **Number** of **subarrays** having even product. 24, Feb 21. Count **subarrays** having product equal to the power of a given Prime **Number** . 22, Mar 21. **Number** of **subarrays** having absolute **sum greater than K** | Set-2. 03, Jul 19. **Numbers** less **than** N which are product of exactly two. Current minimum and maximum **numbers** We would like to show you a description here but the site won't allow us Forth Iterations This means the interviewer will pick a problem with a small concise solution (415) 828-4153 [email protected] We can store the maximum **subarray** **sum** ending at a particular index in an auxiliary array and then traverse the auxiliary array to find the. Given an array of positive integers nums and a positive integer target, return the minimal length of a contiguous subarray [nums l, nums l+1, ..., nums r-1, nums r] of which the **sum** is **greater than** or equal to target. If there is no such subarray, return 0 instead. Example 1:. "/>. Title Description Given an unordered array arr, Where elements can be positive 、 Negativity 、 can 0, Given an integer **k**. seek arr All **subarrays** The cumulative **sum** **of** is less than or equal to **k** The longest **subarray** length of . title The first idea ： Definition dp i For arr i The cumulative **sum** at the end of the position , find dp i Less than equal to **k** The maximum of the values of Can. **Number** **of** **subarrays** having **sum** less than or equal to R - **Number** **of** **subarrays** having **sum** less than or equal to L-1 For finding required **number** **of** **subarrays** refer this: **Subarrays** **with** **sum** less than **k** → Reply. Maximum of all **subarrays** **of** size **k**. Method 1 (Simple): Run two loops to generate all **subarrays** **of** size **k** and find maximum and minimum values. Finally, return **sum** **of** all maximum and minimum elements. Time taken by this solution is O (n*k). Method 2 (Efficient using Dequeue): The idea is to use Dequeue data structure and sliding window concept. Problem Statement: Given an array and a **sum k**, we need to print the length of the longest subarray that sums to **k**. Examples: Example 1: Input: arr = {7,1,6,0}, **k** = 7 Output: Length of the longest subarray **with sum K** is 3 Explanation: 1 + 6 + 0 = 7, it is the longest subarray **with sum** 7 and length 3.Example 2: Input: arr = {2,3,5,1,9}, **k** = 10 Output: Length of the longest. Medium. Given an array of integers arr and two integers **k** and threshold, return the **number** of **sub**-arrays of size **k** and average **greater than** or equal to threshold. Example 1: Input: arr = [2,2,2,2,5,5,5,8], **k** = 3, threshold = 4 Output: 3 Explanation: **Sub**-arrays [2,5,5], [5,5,5] and [5,5,8] have averages 4, 5 and 6 respectively. All other sub.